If x = 5 - 262\sqrt{6}26, find : x2+1x2x^2 + \dfrac{1}{x^2}x2+x21
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Given,
x = 5 - 262\sqrt{6}26
∴1x=15−26\therefore \dfrac{1}{x} = \dfrac{1}{5 - 2\sqrt{6}}∴x1=5−261
Rationalizing,
⇒1x=15−26×5+265+26=5+2652−(26)2=5+2625−24=5+261=5+26.\Rightarrow \dfrac{1}{x} = \dfrac{1}{5 - 2\sqrt{6}} \times \dfrac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} \\[1em] = \dfrac{5 + 2\sqrt{6}}{5^2 - (2\sqrt{6})^2} \\[1em] = \dfrac{5 + 2\sqrt{6}}{25 - 24} \\[1em] = \dfrac{5 + 2\sqrt{6}}{1} \\[1em] = 5 + 2\sqrt{6}.⇒x1=5−261×5+265+26=52−(26)25+26=25−245+26=15+26=5+26.
By formula,
x2+1x2=(x+1x)2−2x^2 + \dfrac{1}{x^2} = \Big(x + \dfrac{1}{x}\Big)^2 - 2x2+x21=(x+x1)2−2
Substituting values we get :
⇒x2+1x2=(5−26+5+26)2−2=102−2=100−2=98.\Rightarrow x^2 + \dfrac{1}{x^2} = (5 - 2\sqrt{6} + 5 + 2\sqrt{6})^2 - 2 \\[1em] = 10^2 - 2 \\[1em] = 100 - 2 \\[1em] = 98.⇒x2+x21=(5−26+5+26)2−2=102−2=100−2=98.
Hence, x2+1x2x^2 + \dfrac{1}{x^2}x2+x21 = 98.
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If x = 23+222\sqrt{3} + 2\sqrt{2}23+22, find :
(i) 1x\dfrac{1}{x}x1
(ii) x+1xx + \dfrac{1}{x}x+x1
(iii) (x+1x)2\Big(x + \dfrac{1}{x}\Big)^2(x+x1)2
If x = 1−21 - \sqrt{2}1−2, find the value of (x−1x)3\Big(x - \dfrac{1}{x}\Big)^3(x−x1)3.
If 2=1.4 and 3\sqrt{2} = 1.4 \text{ and } \sqrt{3}2=1.4 and 3 = 1.7, find the value of each of the following, correct to one decimal place :
(i) 13−2\dfrac{1}{\sqrt{3} - \sqrt{2}}3−21
(ii) 13+22\dfrac{1}{3 + 2\sqrt{2}}3+221
(iii) 2−33\dfrac{2 - \sqrt{3}}{\sqrt{3}}32−3
Evaluate :
4−54+5+4+54−5\dfrac{4 - \sqrt{5}}{4 + \sqrt{5}} + \dfrac{4 + \sqrt{5}}{4 - \sqrt{5}}4+54−5+4−54+5
